Let d≥1 be an integer and let →A:={Ai}di=1 be real numbers. We consider a following integral:
I(d)(→A):=∞∫0e−u2[d∏i=1erf(Aiu)]du
By expanding the error functions in Taylor series and then integrating term by term we found the answer for d=1 and d=2. We have:
√πI(d)(→A)={arctan(A1)if d=1arctan(A1A2√1+A21+A22)if d=2
Now the question is how do we derive the result for arbitrary values of d?
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